name: mike lopez
chapter 2 quiz 2 - 2.4 to 2.6
1. the graduate record examinations (gres) are widely used to help predict the performance of applicants to graduate schools. the scores on the gre chemistry test are approximately normal with mean 694 and standard deviation 112.
a. about what percent of test takers earn a score less than 500 on the gre chemistry test?
b. what proportion of test takers earn a score greater than or equal to 900 on this test?
c. estimate the score at the 99th percentile on the gre chemistry test.
2. bigger animals tend to carry their young longer before birth. the length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days.
a. draw the normal distribution.
b. use the empirical rule to estimate the percentage of horse pregnancies that are longer...

Question

name: mike lopez
chapter 2 quiz 2 - 2.4 to 2.6
1. the graduate record examinations (gres) are widely used to help predict the performance of applicants to graduate schools. the scores on the gre chemistry test are approximately normal with mean 694 and standard deviation 112.
   a. about what percent of test takers earn a score less than 500 on the gre chemistry test?
   b. what proportion of test takers earn a score greater than or equal to 900 on this test?
   c. estimate the score at the 99th percentile on the gre chemistry test.
2. bigger animals tend to carry their young longer before birth. the length of horse pregnancies from conception to birth varies according to a roughly normal distribution with mean 336 days and standard deviation 6 days.
   a. draw the normal distribution.
   b. use the empirical rule to estimate the percentage of horse pregnancies that are longer...

Accepted Answer

### Part 1a: Percent of scores < 500
# Explanation:
## Step1: Calculate z - score
The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 500$, $\mu=694$, and $\sigma = 112$.
So, $z=\frac{500 - 694}{112}=\frac{- 194}{112}\approx - 1.73$.
## Step2: Find the area to the left of z=-1.73
Using the standard normal distribution table (or z - table), the area to the left of $z=-1.73$ is approximately 0.0418 or 4.18%.
# Answer:
Approximately 4.18%

### Part 1b: Proportion of scores ≥ 900
# Explanation:
## Step1: Calculate z - score
Using the z - score formula $z=\frac{x-\mu}{\sigma}$, with $x = 900$, $\mu = 694$, and $\sigma=112$.
$z=\frac{900 - 694}{112}=\frac{206}{112}\approx1.84$.
## Step2: Find the area to the right of z = 1.84
The area to the left of $z = 1.84$ from the z - table is approximately 0.9671. So the area to the right (scores ≥ 900) is $1 - 0.9671=0.0329$ or 3.29%.
# Answer:
Approximately 0.0329 (or 3.29%)

### Part 1c: 99th percentile score
# Explanation:
## Step1: Find the z - score for 99th percentile
The 99th percentile means that 99% of the data is below this value. Looking at the z - table, the z - score corresponding to an area of 0.99 is approximately 2.33 (more accurately, we can use a z - table or calculator for more precision).
## Step2: Use the z - score formula to find x
The z - score formula is $z=\frac{x-\mu}{\sigma}$, we can rearrange it to $x=\mu+z\sigma$.
Given $\mu = 694$, $\sigma = 112$, and $z\approx2.33$.
$x=694+2.33	imes112=694 + 260.96=954.96\approx955$.
# Answer:
Approximately 955

### Part 2a: Drawing the normal distribution for horse pregnancies
# Explanation:
## Step1: Identify the mean and standard deviation
The mean ($\mu$) of horse pregnancies is 336 days and the standard deviation ($\sigma$) is 6 days.
## Step2: Mark the values on the horizontal axis
For a normal distribution, we mark the mean in the center. Then we mark values at $\mu-\sigma=336 - 6 = 330$, $\mu - 2\sigma=336-12 = 324$, $\mu-3\sigma=336 - 18=318$, $\mu+\sigma=336 + 6=342$, $\mu + 2\sigma=336+12 = 348$, $\mu+3\sigma=336 + 18=354$.
## Step3: Draw the normal curve
Draw a bell - shaped curve centered at 336. The curve should be symmetric about the mean, with the highest point at the mean, and tapering off as we move away from the mean in both directions.
# Answer:
The normal distribution is drawn with the mean (336) in the center, and values 318, 324, 330, 336, 342, 348, 354 marked on the horizontal axis, with a symmetric bell - shaped curve.

### Part 2b: Percentage of horse pregnancies longer than... (assuming we want to find the percentage longer than, say, 342 days, but from the writing it seems related to the empirical rule)
# Explanation:
## Step1: Recall the empirical rule
The empirical rule states that for a normal distribution:
 - Approximately 68% of the data lies within $\mu\pm\sigma$ (336 ± 6, i.e., 330 - 342).
 - Approximately 95% of the data lies within $\mu\pm2\sigma$ (336 ± 12, i.e., 324 - 348).
 - Approximately 99.7% of the data lies within $\mu\pm3\sigma$ (336 ± 18, i.e., 318 - 354).

If we want to find the percentage of pregnancies longer than 342 days (which is $\mu+\sigma$), the percentage of data above $\mu+\sigma$ is $\frac{100 - 68}{2}=16\%$ (since the data is symmetric, the area above $\mu+\sigma$ is half of the area outside $\mu\pm\sigma$).
# Answer:
Approximately 16% (assuming the question is about pregnancies longer than 342 days, consistent with the written $100 - 84 = 16\%$)

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QUESTION IMAGE

Question

Explanation:

Part 1a: Percent of scores < 500

Step1: Calculate z - score

Step2: Find the area to the left of z=-1.73

Step1: Calculate z - score

Step2: Find the area to the right of z = 1.84

Step1: Find the z - score for 99th percentile

Step2: Use the z - score formula to find x

Answer:

Part 1b: Proportion of scores ≥ 900